0.00/0.26	YES
0.00/0.26	
0.00/0.26	DP problem for innermost termination.
0.00/0.26	P =
0.00/0.26	  eval#(x, y) -> eval#(x, y) [x + y > 0 && 0 >= x && 0 >= y]
0.00/0.26	  eval#(I0, I1) -> eval#(I0, I1 - 1) [I0 + I1 > 0 && 0 >= I0 && I1 > 0]
0.00/0.26	  eval#(I2, I3) -> eval#(I2 - 1, I3) [I2 + I3 > 0 && I2 > 0]
0.00/0.26	R = 
0.00/0.26	  eval(x, y) -> eval(x, y) [x + y > 0 && 0 >= x && 0 >= y]
0.00/0.26	  eval(I0, I1) -> eval(I0, I1 - 1) [I0 + I1 > 0 && 0 >= I0 && I1 > 0]
0.00/0.26	  eval(I2, I3) -> eval(I2 - 1, I3) [I2 + I3 > 0 && I2 > 0]
0.00/0.26	
0.00/0.26	The dependency graph for this problem is:
0.00/0.26	  0 -> 
0.00/0.26	  1 -> 1
0.00/0.26	  2 -> 1, 2
0.00/0.26	Where:
0.00/0.26	  0) eval#(x, y) -> eval#(x, y) [x + y > 0 && 0 >= x && 0 >= y]
0.00/0.26	  1) eval#(I0, I1) -> eval#(I0, I1 - 1) [I0 + I1 > 0 && 0 >= I0 && I1 > 0]
0.00/0.26	  2) eval#(I2, I3) -> eval#(I2 - 1, I3) [I2 + I3 > 0 && I2 > 0]
0.00/0.26	
0.00/0.26	We have the following SCCs.
0.00/0.26	  { 2 }
0.00/0.26	  { 1 }
0.00/0.26	
0.00/0.26	DP problem for innermost termination.
0.00/0.26	P =
0.00/0.26	  eval#(I0, I1) -> eval#(I0, I1 - 1) [I0 + I1 > 0 && 0 >= I0 && I1 > 0]
0.00/0.26	R = 
0.00/0.26	  eval(x, y) -> eval(x, y) [x + y > 0 && 0 >= x && 0 >= y]
0.00/0.26	  eval(I0, I1) -> eval(I0, I1 - 1) [I0 + I1 > 0 && 0 >= I0 && I1 > 0]
0.00/0.26	  eval(I2, I3) -> eval(I2 - 1, I3) [I2 + I3 > 0 && I2 > 0]
0.00/0.26	
0.00/0.26	We use the reverse value criterion with the projection function NU:
0.00/0.26	  NU[eval#(z1,z2)] = z1 + z2 + -1 * 0
0.00/0.26	
0.00/0.26	This gives the following inequalities:
0.00/0.26	  I0 + I1 > 0 && 0 >= I0 && I1 > 0  ==>  I0 + I1 + -1 * 0 > I0 + (I1 - 1) + -1 * 0  with  I0 + I1 + -1 * 0 >= 0
0.00/0.26	
0.00/0.26	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/0.26	
0.00/0.26	DP problem for innermost termination.
0.00/0.26	P =
0.00/0.26	  eval#(I2, I3) -> eval#(I2 - 1, I3) [I2 + I3 > 0 && I2 > 0]
0.00/0.26	R = 
0.00/0.26	  eval(x, y) -> eval(x, y) [x + y > 0 && 0 >= x && 0 >= y]
0.00/0.26	  eval(I0, I1) -> eval(I0, I1 - 1) [I0 + I1 > 0 && 0 >= I0 && I1 > 0]
0.00/0.26	  eval(I2, I3) -> eval(I2 - 1, I3) [I2 + I3 > 0 && I2 > 0]
0.00/0.26	
0.00/0.26	We use the reverse value criterion with the projection function NU:
0.00/0.26	  NU[eval#(z1,z2)] = z1
0.00/0.26	
0.00/0.26	This gives the following inequalities:
0.00/0.26	  I2 + I3 > 0 && I2 > 0  ==>  I2 > I2 - 1  with  I2 >= 0
0.00/0.26	
0.00/0.26	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.24	EOF